(i) Express \(f(x)\) in partial fractions:
Assume \(\frac{5x^2 + x + 27}{(2x + 1)(x^2 + 9)} = \frac{A}{2x+1} + \frac{Bx + C}{x^2 + 9}\).
Multiply through by the denominator: \(5x^2 + x + 27 = A(x^2 + 9) + (Bx + C)(2x + 1)\).
Expand and equate coefficients:
\(5x^2 + x + 27 = Ax^2 + 9A + 2Bx^2 + Bx + Cx + C\).
\(5x^2 + x + 27 = (A + 2B)x^2 + (B + C)x + (9A + C)\).
Equating coefficients gives:
\(A + 2B = 5\)
\(B + C = 1\)
\(9A + C = 27\)
Solving these equations, we find \(A = 3\), \(B = 1\), \(C = 0\).
Thus, \(f(x) = \frac{3}{2x+1} + \frac{x}{x^2 + 9}\).
(ii) Integrate \(f(x)\):
\(\int_0^4 f(x) \, dx = \int_0^4 \left( \frac{3}{2x+1} + \frac{x}{x^2 + 9} \right) \, dx\).
Integrate each term separately:
\(\int \frac{3}{2x+1} \, dx = \frac{3}{2} \ln |2x+1|\).
\(\int \frac{x}{x^2 + 9} \, dx = \frac{1}{2} \ln |x^2 + 9|\).
Evaluate from 0 to 4:
\(\left[ \frac{3}{2} \ln |2x+1| \right]_0^4 + \left[ \frac{1}{2} \ln |x^2 + 9| \right]_0^4\).
\(= \frac{3}{2} (\ln 9 - \ln 1) + \frac{1}{2} (\ln 25 - \ln 9)\).
\(= \frac{3}{2} \ln 9 + \frac{1}{2} \ln \frac{25}{9}\).
\(= \ln 9^{3/2} + \ln \left( \frac{25}{9} \right)^{1/2}\).
\(= \ln 27 + \ln \frac{5}{3}\).
\(= \ln (27 \times \frac{5}{3})\).
\(= \ln 45\).
